A pole stands vertically inside a triangular park ABC. Let the angle of elevation of the top of the

Q: Why is $D$ the circumcenter? Can't it be the incenter or centroid?

Great question! The key lies in the equal angles of elevation. Because the height of the pole $h$ is constant and the angle of elevation $\theta = \frac{\pi}{3}$ is constant for all three vertices, the base distance $AD = BD = CD = \frac{h}{\tan(\theta)}$ must be identical. By definition, the only point in a triangle that is equidistant from all three vertices is the circumcenter. The incenter is equidistant from the sides, and the centroid is the intersection of medians, so neither fits this specific geometric condition.

Q: Do I need to know the coordinates of $A, B,$ and $C$ to solve this?

Not at all! This is the beauty of geometry. While you could set up a coordinate system, it would be unnecessarily complex. The problem provides the circumradius $R$ directly, which is the only piece of information needed to link the triangle's geometry to the pole's height. Trust the properties of the circumcenter; they are designed to simplify your life!

Q: Is the pole necessarily inside the triangle? What if the triangle is obtuse?

Excellent observation. If the triangle is obtuse, the circumcenter actually lies outside the triangle. However, the problem states the pole stands 'inside' the park. This implies the triangle is acute or right-angled, ensuring the circumcenter is within or on the boundary of the park. Regardless of the triangle's type, the relationship $AD = R$ holds true, and the math remains consistent.

Question

A pole stands vertically inside a triangular park $ABC$. Let the angle of elevation of the top of the pole from each corner of the park be $\frac{\pi}{3}$. If the radius of the circumcircle of $\Delta ABC$ is 2, then the height of the pole is equal to:

Options:

(A) $\frac{2\sqrt{3}}{3}$

(B) $2\sqrt{3}$

(C) $\sqrt{3}$

(D) $\frac{1}{\sqrt{3}}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a perfectly triangular park, denoted by vertices $A$, $B$, and $C$. The ground is flat, and at some point $D$ inside this park, a vertical pole of height $h$ stands tall, with its top at point $P$.

This is a beautiful exercise in visualizing three-dimensional space projected onto a two-dimensional plane. When you look up at the top of the pole $P$ from any corner of the park, you create a line of sight. The problem states that the angle of elevation from each corner is $\frac{\pi}{3}$.

### The Right-Angled Revelation

Let us focus on the triangles formed by the pole and the ground. We have three right-angled triangles: $\Delta PDA$, $\Delta PDB$, and $\Delta PDC$.

In each of these, the pole $PD$ is the perpendicular side of length $h$, and the segments $AD$, $BD$, and $CD$ are the bases on the ground. Because the angle of elevation is $\frac{\pi}{3}$ for all three, we use the trigonometric relationship:

$$	an\left(\frac{\pi}{3}ight) = \frac{PD}{AD}$$

Since $	an\left(\frac{\pi}{3}ight) = \sqrt{3}$, we have $\sqrt{3} = \frac{h}{AD}$, which implies $AD = \frac{h}{\sqrt{3}}$. Because this logic applies identically to $BD$ and $CD$, we conclude:

$$AD = BD = CD = \frac{h}{\sqrt{3}}$$

### The Circumcenter Connection

Here is where the geometry reveals its structure. We have found a point $D$ on the ground that is equidistant from all three vertices of the triangle $ABC$.

In geometry, a point equidistant from the vertices of a triangle is the **circumcenter**. Therefore, $D$ is the circumcenter of $\Delta ABC$.

The distance from the circumcenter to any vertex is defined as the circumradius, denoted by $R$. The problem explicitly gives us $R = 2$. Thus, we have the equality:

$$AD = R = 2$$

### The Final Synthesis

Now, we bring our pieces together. We know $AD = 2$, and we previously established that $AD = \frac{h}{\sqrt{3}}$.

Setting these equal, we get:

$$\frac{h}{\sqrt{3}} = 2$$

Solving for $h$, we find:

$$h = 2\sqrt{3}$$

The height of the pole is **$2\sqrt{3}$**. In JEE Advanced, the most difficult problems often yield to the most elegant geometric insights. Keep visualizing, keep questioning, and keep pushing forward!

Select Answer:

Visualized Solution

Visualizing the Setup

The Vertical Pole

Lines of Sight

Angle of Elevation

Forming Right Triangles

Trigonometric Ratio

Equating the Bases

The Circumcenter

Using the Circumradius

Substituting Values

Evaluating the Tangent

Final Calculation

The Sigma Insight: Heights and Distances

Analyzing the Setup

The Right-Angled Revelation

The Circumcenter Connection

The Final Synthesis

Similar Questions

A vertical pole stands at a point $Q$ on a horizontal ground. $A$ and $B$ are points on the ground, $d$ meters apart. The pole subtends angles $α$ and $β$ at $A$ and $B$ respectively. $A B$ subtends an angle $γ$ at $Q$ . Find the height of the pole.

ABC is a triangular park with AB = AC = 100 metres. A vertical tower is situated at the mid-point of BC. If the angles of elevation of the top of the tower at A and B are $cot^{- 1} (32)$ and $csc^{- 1} (22)$ respectively, then the height of the tower (in metres) is :

A tower stands at the centre of a circular park. $A$ and $B$ are two points on the boundary of the park such that $A B (= a)$ subtends an angle of $6 0^{\circ}$ at the foot of the tower, and the angle of elevation of the top of the tower from $A$ or $B$ is $3 0^{\circ}$ . The height of the tower is

From the base of a pole of height 20 meter, the angle of elevation of the top of a tower is $6 0^{\circ}$ . The pole subtends an angle $3 0^{\circ}$ at the top of the tower. Then the height of the tower is:

Consider a triangular plot ABC with sides AB=7m, BC=5m and CA=6m. A vertical lamp-post at the mid point D of AC subtends an angle $3 0^{\circ}$ at B. The height (in m) of the lamp-post is:

$A B C$ is a triangular park with $A B = A C = 100$ m. A television tower stands at the midpoint of $B C$ . The angles of elevation of the top of the tower at $A, B, C$ are $4 5^{\circ}, 6 0^{\circ}, 6 0^{\circ}$ , respectively. Find the height of the tower.

A pole stands vertically inside a triangular park $Δ A B C$ . If the angle of elevation of the top of the pole from each corner of the park is same, then in $Δ A B C$ the foot of the pole is at the

A vertical pole fixed to the horizontal ground is divided in the ratio $3 : 7$ by a mark on it with lower part shorter than the upper part. If the two parts subtend equal angles at a point on the ground $18 m$ away from the base of the pole, then the height of the pole (in meters) is :

The upper $\frac{3}{4}$ th portion of a vertical pole subtends an angle $tan^{- 1} \frac{3}{5}$ at a point in the horizontal plane through its foot and at a distance 40 m from the foot. A possible height of the vertical pole is